ISE313 COMPUTER INTEGRATED MANUFACTURING AND AUTOMATION I
FINAL EXAM15.01.2008
1. Name the three cases of part or product variety in manufacturing systems. Briefly define each of the three cases. (10 points)
Answer: The three cases of part or product variety in manufacturing systems are
(1) single model,
(2) batch model, and
(3) mixed model.
In the single-model case, all parts or products made by the manufacturing system are identical. In the batch-model case, different parts or products are made by the system, but they are made in batches because the physical setup and/or equipment programming must be changed over between models. In the mixed-model case, different parts or products are made by the system, but the differences are not significant, so the system is able to handle them without the need for time-consuming changeovers in setup or program.
2. What is a machine cluster? (5 points)
Answer: As defined in the text, a machine cluster is a collection of two or more machines producing parts or products with identical cycle times and serviced (usually loaded and unloaded) by one worker.
3. What are the five enablers that are required for unattended operation of a single-model or batch-model automated production cell? (10 points)
Answer: As given in the text, the five enablers of unattended operation for a single-model or batch-model automated production cell are the following:
(1) a programmed cycle that allows the machine to perform every step of the processing or assembly cycle automatically;
(2) a parts storage subsystem and a supply of parts that permit continuous operation beyond one machine cycle;
(3) automatic transfer of workparts between the storage system and the machine (automatic unloading of finished parts from the machine and loading of raw workparts to the machine);
(4) periodic attention of a worker who performs the necessary machine tending functions (e.g., parts loading and unloading of the storage subsystem) for the particular processing or assembly operation; and
(5) built-in safeguards that protect the system against operating under conditions that may be unsafe or self-destructive to itself or destructive to the work units being processed or assembled.
4. A worker is currently responsible for tending two machines in a machine cluster. The service time per machine is 0.35 min and the time to walk between machines is 0.15 min. The machine automatic cycle time is 1.90 min. If the worker's hourly rate = $12/hr and the hourly rate for each machine = $18/hr, determine (a) the current hourly rate for the cluster, and (b) the current cost per unit of product, given that two units are produced by each machine during each machine cycle. (c) What is the % idle time of the worker? (d) What is the optimum number of machines that should be used in the machine cluster, if minimum cost per unit of product is the decision criterion? (20 points)
Solution: (a) Co = $12 + 2($18) = $48.00/hr
(b) Tc = Tm + Ts = 1.90 + 0.35 = 2.25 min/cycle
Rc = 2(2) = 106.67 pc/hrCpc = = $0.45/pc
(c) Worker engagement time/cycle = 2(Ts + Tr) = 2(0.35 + 0.15) = 1.0 min
Idle time IT = = 0.555 = 55.5%
(d) n = = 2.25/0.5 = 4.5 machines
n1 = 4 machines: Cpc(4) = 0.5(12/4 + 18(2.25/60) = $0.394/pc
n2 = 5 machines: Cpc(5) = 0.5(12 + 18x5)(0.50/60) = $0.425/pc
Use n1 = 4 machines
5. Name four of the six measures used to assess the performance of a storage system? (10 points)
Answer: The six performance measures discussed in the text are the following:
(1) storage capacity, which is defined and measured either as the total volumetric space available or as the total number of storage compartments in the system available for items or loads;
(2) storage density, defined as the volumetric space available for actual storage relative to the total volumetric space in the storage facility;
(3) accessibility, which refers to the capability to access any desired item or load stored in the system;
(4) system throughput, defined as the hourly rate at which the storage system receives and puts loads into storage and/or retrieves and delivers loads to the output station;
(5) utilization, which is defined as the proportion of time that the system is actually being used for performing storage and retrieval operations compared with the time it is available, and
(6) availability, defined as the proportion of time that the system is capable of operating (not broken down) compared with the normally scheduled shift hours.
6. An AS/RS with four aisles is 80 m long and 18 m high. The S/R machine has a maximum speed of 1.6 m/s in the horizontal direction. It accelerates from zero to 1.6 m/s in a distance of 2.0 m. On approaching its target position (where the S/R machine will transfer a load onto or off of its platform), it decelerates from 1.6 m/s to a full stop in 2.0 m. The maximum vertical speed is 0.5 m/s, and the acceleration and deceleration distances are each 0.3 m. Rates of acceleration and deceleration are constant in both directions. Pick and deposit time = 12 s. Utilization of the AS/RS is assumed to be 90%, and the number of dual command cycles = the number of single command cycles. (a) Calculate the single command and dual command cycle times, including considerations for acceleration and deceleration. (b) Determine the throughput rate for the system. (25 points)
Solution: (a) Horizontal travel: v = = at = 1.6 m/s,a = 1.6/t
y = = at2/2 = 0.5(1.6/t) t2 = 0.8 t = 2 mt = 2/0.8 = 2.5 s.
Vertical travel: v = = at = 0.5 m/s,a = 0.5/t
z = = at2/2 = 0.5(0.5/t) t2 = 0.25 t = 0.3 mt = 0.3/0.25 = 1.2 s.
Tcs = 2 Max+ 2(12) = 79 s/cycle = 1.3167 min/cycle
Tcd = 2 Max+ 4(12) = 130.5 s/cycle = 2.175 min/cycle
(b) Rcs Tcs + Rcd Tcd = 60(0.90) = 54 min
Given Rcs = Rcd, 1.3167 Rcs + 2.175 Rcs = 3.4917 Rcs = 54.0Rcs = 15.465 cycles/hr
Rcd = Rcs = 15.465 cycles/hr
Rt = Rcs + 2 Rcd = 15.465 + 2(15.465) = 46.396 transactions/hr per aisle
With 4 aisles, Rt = 4(46.396) = 185.6 transactions/hr
7. A recirculating conveyor has a total length of 700 ft and a speed of 90 ft/min. Spacing of part carriers = 14 ft. Each carrier can hold one part. Automatic machines load and unload the conveyor at the load and unload stations. Time to load a part is 0.10 min and unload time is the same. To satisfy production requirements, the loading and unloading rates are each 2.0 parts per min. Evaluate the conveyor system design with respect to the (i) speed rule, (ii) capacity constraint, and (iii) uniformity principle. (20 points)
Solution: (1) Speed rule: Lower limit: Max{RL, Ru}
= 6.428 parts/min Max{2, 2} (OK)
Upper limit: Min{}
= 6.428 carrier/min Min{} = Min{10, 10} (OK)
(2) Capacity constraint: In this case, the flow rate is interpreted to be the specified load and unload rates,
that is, Rf = RL = Ru = 2 parts/min
= 6.428 parts/min 2 (OK)
(3) Uniformity principle: The conveyor is assumed to be uniformly loaded throughout its length since
RL = Ru and the flow rate capacity is significantly greater than RL and Ru. (OK)
Conclusion: The system is feasible.